In the early 20th century there was a lot of fuss over the axiom of choice implying that there are Lebesgue non-measurable sets of reals. In his book about The Axiom of Choice, Gregory Moore points to the following paper:
Sierpinski, W. "L’axiome de M. Zermelo et son rôle dans la théorie des ensembles et l’analyse." Bulletin de l’Académie des Sciences de Cracovie, Classe des Sciences Math., Sér. A (1918) (1918): 97-152.
(And even more specifically, to pages 124-125.)
Moore writes that Sierpinski proved that if $[\Bbb R]^\omega$ has cardinality $2^{\aleph_0}$, then there is a non-measurable set. Lebesgue argued that Sierpinski uses the axiom of choice, but Moore points out that while the assumption requires some amount of choice (as we full well know today), the implication does not.
Being a historical book about the axiom of choice, Moore doesn't provide a sketch of the proof. However, I've been unable to locate the paper. Which brings me to my question.
Question. Is there any accessible (preferably English) reference to what the argument of Sierpinski was?
Best Answer
Here's Sierpinski's argument: Let $h:[\mathbb{R}]^{\omega} \to \mathbb{R}$ be any injection. Define $f:\mathbb{R} \to \mathbb{R}$ by $f(x) = h(E_x)$ where $E_x$ is the set of all reals which are at a rational distance from $x$. Note that $x - y$ is rational iff $f(x) = f(y)$. Towards a contradiction, suppose $f$ is Lebesgue measurable. Then $g(x) = f(x) - f(-x)$ is also measurable and is nonzero precisely at irrationals. Let $N = \{x: g(x) > 0\}$. Note that for any rational $r$, $g(r - x) = -g(x)$ so $x \in N \iff r - x \notin N$. But this contradicts Lebesgue density theorem.